Codeforces Global Round 1 E. Magic Stones

链接

http://codeforces.com/contest/1110/problem/E

题解

观察式子
c i ′ = c i + 1 + c i − 1 − c i c_i'=c_{i+1}+c_{i-1}-c_i ci=ci+1+ci1ci
其实就是
c i ′ − c i − 1 = c i + 1 − c i c_i'-c_{i-1}=c_{i+1}-c_i cici1=ci+1ci

c i + 1 − c i ′ = c i − c i − 1 c_{i+1}-c_{i}'=c_{i}-c_{i-1} ci+1ci=cici1
其实就是在差分序列种交换了第 i − 1 i-1 i1项和第 i i i
那我就看下这两个序列是不是首项相等而且差分序列排序之后相等就好了

代码

#include 
#define cl(x) memset(x,0,sizeof(x))
#define maxn 100010
#define linf (1ll<<60)
#define iinf 0x3f3f3f3f
#define dinf 1e100
#define eps 1e-8 
using namespace std;
typedef long long ll;
ll N, M, buf[100], c[maxn], t[maxn], d1[maxn], d2[maxn];
ll read(ll x=0)
{
	ll c, f=1;
	for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
	for(;isdigit(c);c=getchar())x=x*10+c-48;
	return f*x;
}
void init()
{
	ll i;
	N = read();
	for(i=1;i<=N;i++)c[i] = read();
	for(i=1;i<=N;i++)t[i] = read();
	for(i=1;i<=N;i++) d1[i]=c[i]-c[i-1], d2[i]=t[i]-t[i-1];
	sort(d1+2,d1+N+1);
	sort(d2+2,d2+N+1);
}
bool check()
{
	ll i;
	for(i=2;i<=N;i++)if(d1[i]!=d2[i])return false;
	return c[1]==t[1];
}
int main()
{
	init();
	if(check())printf("Yes");
	else printf("No");
	return 0;
}

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